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3 years ago

5

In this version of golf the winner is the golfer who wins the greater number of holes. 1.miniature golf

2 answers:

Veseljchak [2.6K]3 years ago

8 0

It’s gonna be tournament play

professor190 [17]3 years ago

8 0

Answer:

Tournament play

Explanation:

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Children are ready for more complex storybooks:

FromTheMoon [43]

Answer:

4

Explanation:

Because that is when they're mind learns how to read

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3 years ago

Sam has been working to improve his muscular fitness. He jumps rope and trains with weights. What will most likely be the result

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Increased Bone Density

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4 years ago

Read 2 more answers

Gold has a density of 19.3 grams per cubic centimeter. what is the density of gold in metric tons per cubic meter?

valina [46]

In order to determine this, we will first need some conversions. We will need to convert metric tons and grams into one another and also cubic centimeters to cubic meters into one another.

1 metric ton = 1000 kg

1 kg = 1000 grams

1 metric ton = 10⁶ grams

So 10⁶ grams / metric ton

1 meter = 100 cm
1 m³ = (100)³ cm³
1 m³ = 10⁶ cm³

So 10⁶ cm⁶ / m³

Now, we manipulate the given value:

(19.3 grams / cm³) * (1 metric ton / 10⁶ grams) * (10⁶ cm³ / m³)
= 19.3 metric tons / m³

The density of gold is 19.3 metric tons meter meter cubed.

6 0

3 years ago

How are electrons distributed in the molecular orbitals, when the 1s orbitals of two hydrogen atoms combine to form a hydrogen m

Evgesh-ka [11]

Answer:

2 Electrons

Explanation:

Both the Hydrogen atom will share electron equally by covalent bonding. so there will be 2 electron in the bonding molecular orbital.

3 0

3 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?

SpyIntel [72]

Answer:

U= $0.112 x10^{-6}$ J

u= 0.708 $\frac{J}{m^{3} }$

Explanation:

diameter= 2 cm

$r= 1 cm * \frac{1m}{100cm} = 0.01 m$

distance= 0.5 mm

$d= 0.5 mm * \frac{1m}{1000m}= 0.5 x10^{-3}$

Area A= $\pi *r^{2}= 0.314x10^{-3} m^{2}$

Volume v= $A*d = 0.314 x10^{-3}m^{2}*0.5x10^{-3} m =0.157x10^{-6} m^{3}$

$v= 0.157 x10^{-6} m^{3}$

Constant vacuum permittivity

$E_{o}= 8.85x10^{-12}$

a).

$C= \frac{A*E_{o} }{d} = \frac{0.314x10^{-3} *8.85x10^{-12} }{0.5x10^{-3} } \\C= 5.56 x10^{-12}$ F

$U= \frac{1}{2} *C *V^{2}\\ U=\frac{1}{2} * 5.56x10^{-12}*(200)^{2} \\U=0.112 x^{-6} J\\$

b).

$u=\frac{U}{v}$

$u=\frac{0.112 x10^{-6}J}{0.157x10^{-6}m^{3} } \\u=0.708 \frac{J}{m^{3} }$

5 0

4 years ago